A 512cm\(^3\) sample of a gas weighed 1.236g at 20\(^{\circ}\)C and a pressure of one...

Explanation

To determine the relative molecular mass of the gas, we need to use the ideal gas equation:

\(PV = nRT\)

where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount of substance in moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

First, convert the temperature to Kelvin:

\(T = 20^{\circ}C + 273.15 = 293.15K\)

Next, convert the volume to meters cubed:

\(V = 512 cm^3 = 0.000512 m^3\)

Now, we can rearrange the ideal gas equation to solve for the amount of substance, \(n\):

\(n = \frac{PV}{RT}\)

Substitute the given values:

\(n = \frac{(1 atm)(0.000512 m^3)}{(8.314 JK^{-1}mol^{-1})(293.15K)}\)

Convert the pressure from atm to J m\(^{-3}\):

\(1 atm = 101,325 J m^{-3}\)

\(n = \frac{(101,325 J m^{-3})(0.000512 m^3)}{(8.314 JK^{-1}mol^{-1})(293.15K)}\)

Calculate \(n\):

\(n \approx 0.0213 mol\)

Now, we can determine the relative molecular mass by dividing the mass of the gas by the amount of substance:

Relative Molecular Mass = \(\frac{mass}{moles}\)

Relative Molecular Mass = \(\frac{1.236g}{0.0213 mol} \approx 58.07\)

So, the correct answer isOption A: 58.07.